Find the center of the symmetry group $S_n$. 
Find the center of the symmetry group $S_n$.

Attempt:
By definition, the center is $Z(S_n) = \{ a \in S_n : ag = ga \forall\  g \in S_n\}$. Then we know the identity $e$ is in $S_n$ since there is always the trivial permutation. Suppose $a$ is in $S_n$, but not equal to identity. Now we can imagine the permutation as bijective function that maps from $\{1,2,\dotsc,n\}$ to $\{1,2,\dotsc,n\}$.
So suppose $p$ is a permutation map. Then $p$ maps from a location $i$ to a location $j$. Take $p(i) = j$ where $i\neq j$. Let $k$ be in $\{1,2,\dotsc,n\}$, where $k$, $i$ and $j$ are all different elements. The cycle $r = (jk)$, then we will see if this commutes. 
$rp(i) = rj$  
Can someone please help me, I am stuck?  Thank you.
 A: Let $h $ be in the center. Then $gh = hg $ for all $g ∈ S_n$. Let $g \in G_x,$ the
stabilizer of $x \in X$ (we realize $S_n$ as the group of permutations on a finite set $X$ with n elements). Then $g · hx = hx$ for all $g \in G_x$. So $g \in G_{hx}$, as well. This means that $G_x \le G_{hx}$ for all $x \in  X$. Suppose $hx \neq x$ for some $x \in X$. Then every permutation of $X$ that fixes $x$ must also fix $hx$. Since $n > 3$, let $x, hx, y \in X,$ all
distinct. Define $\sigma : X \to X$ by the following:
$x \to x$
$hx \to y$
$y \to hx,$
and $\sigma$ fixes everything else. Then $\sigma$ is a permutation of $X$ that fixes $x$ but does not fix $hx$. This is a contradiction. Hence it must not be the case that $hx \ne x$ for some
$x \in X$, that is, $hx = x$ for all $x \in X$. So the center is trivial
A: Here is a different proof based on the fact that the center of any group is precisely the set of elements whose conjugacy classes are singletons.
For $S_n$, the conjugacy classes are in bijection with partitions of $n$ (since conj. classes are determined by cycle type.) We'll focus on $n\geq 3$ since $S_2$ is abelian. So fix an arbitrary non-increasing partition $\lambda_1,\cdots,\lambda_k$ of $n$ whose largest part is $>1$ (otherwise it's the conj. class of the identity).
The conjugacy class of $\sigma = (1\cdots\lambda_1)(\lambda_1{+}1\ \cdots\ \lambda_1{+}\lambda_2)\cdots((\sum_{i=1}^{k-1}\lambda_i){+}1\ \cdots\ n)$ has an element distinct from $\sigma$ since we can switch 1 & 3 (remember $\lambda_1>1$ and $n\geq 3$). Thus there are no nontrivial conjugacy classes with only 1 element for $n\geq 3$, so the center of $S_n$, $n\geq 3$, is trivial.
A: If $n=2$; $S_2$ is cyclic of order $2$, so it is abelian and $Z(S_2)=S_2$. 
Suppose $n>2$. If $\sigma \in S_n$ is not the identity, then it moves at least one letter $i$, say $\sigma(i)=k$ and since $i\neq k$, it also moves $k$, say $\sigma(k)=j$. Can you produce a permutation (a simple one, don't think too hard) that doesn't commute with $\sigma$? 
Spoiler For example, say $\sigma(i)=k$ and $\sigma(k)=i$; (so $i=j$), and $\sigma$ is of the form $\sigma=(ij)\tau$, with $\tau(ij)=\tau(ji)$. Note that $\tau$ fixes $i,j$, and cannot map something to $i$ or $j$. Then consider $(i\ell)$, a transposition. Then $\sigma(i\ell)$ doesn't map $i$ to $j$: if $\tau$ moves $\ell$, it moves $i$ to something different from $j$; and if $\tau$ doesn't move $\ell$, $i\to\ell$ -- but $(i\ell)\sigma$  maps $i\to j$, so $(i\ell)\sigma\neq \sigma(i\ell)$.
It remains you see what happens when $i\neq j$; but it shouldn't be too hard either.
A: Suppose that $\sigma$ is a permutation of $\{1,\dots,n\}$, where $n\ge 3$. If $\sigma$ commutes with $(1 \; 2)$, then either $1$ and $2$ are both fixed points , or $(1 \; 2)$ is a cycle. Similarly, if $\sigma$ commutes with the transposition $(2 \; 3)$, then either $2$ and $3$ are both fixed points, or $(2 \; 3)$ is a cycle. If $\sigma$ commutes with both $(1 \; 2)$ and with $(2 \; 3)$, it must therefore be the case that $1$, $2$, and $3$ are all fixed points. Inductively, this means that if $\sigma$ commutes with $(1 \; 2),(2\; 3),\dots,(n-1\; n)$, then $1,\dots, n$ are all fixed points, i.e. $\sigma$ is the identity.
